I fly a lot. Talks, meetings, whatever. I usually prefer an aisle seat, because then the rude guy who smells funny and spreads over 1.8 seats only irritates me on one side, and I’m not wedged up against the window.

However, sometimes I do like to grab a window seat, especially if I’m flying near sunset, or over a particularly interesting landscape (flying over southern Utah near sunset will change your life). But even then, the landscape blows past, and eventually you wind up flying over eastern Colorado, and there’s nothing to see but flat, flat land, extending all the way to the horizon.

And as I gaze over the amber waves of grain to the line that divides land and sky, I sometimes wonder how far away that line is. The horizon is a semi-mythical distance, used in poetry as a metaphor for a philosophical division of some kind. But in fact it’s a real thing, and the distance to it can be determined. All it takes is a little knowledge of geometry, and a diagram to show you the way.

Follow along with me here. We’re going to find the lost horizon.

So you’re standing on the Earth. Let’s assume the Earth is a perfect sphere, because that makes things a lot easier. What does our situation look like? Well, it looks something like this:

In this diagram, the circle is the surface of the Earth, which has a radius of R. The Earth’s radius varies with latitude, but I’ll just use 6365 kilometers as a decent average. The dude standing on the Earth is a human of height h (not to scale, huge duh there). The line-of-sight to the horizon is the red line, labeled d. Finding the value of d is the goal here. Note that the radius of the Earth is a constant, but that d will vary as h goes up or down.

The key thing here is that at the visible horizon, the angle between your line-of-sight and the radius line of the Earth is a right angle (marked in the diagram). That means we have a right triangle, and — reach back into the dim, dusty memory of high school — that means we can use the Pythagorean Theorem to get d. The square of the hypotenuse is equal to the sum of the squares of the other two sides. One side is d, the other is R, and the hypotenuse is the Earth’s radius plus your height above the surface, R+h. This gives us the following algebraic formula:

Hey, we have a factor of R2 on both sides, so they cancel! That leaves us with:

d2 = h2 + 2Rh

Now, take the square root of both sides, and voila! You get d.

So now we have an equation that tells us how far away the horizon is depending on where we are above the surface. We can use this to put in different values for h, our height, and see how far away the edge of the Earth is. I put this into an Excel spreadsheet, and the numbers are below.

In the table, the first column is your height in meters above the Earth’s surface (really the height of your eyes) and the second column is the horizon distance in kilometers. Columns three and four are the same, but in feet and miles for you Amurcans.

Height (meters)

Distance (km)

Height(feet)

Distance (miles)

0

0.0

0.0

0.0

1

3.6

3.3

2.1

2

5.1

6.6

3.0

3

6.2

9.8

3.7

4

7.1

13.1

4.3

5

8.0

16.4

4.8

6

8.7

19.7

5.2

7

9.4

23.0

5.7

8

10.1

26.2

6.1

9

10.7

29.5

6.4

10

11.3

32.8

6.8

20

16.0

65.6

9.6

30

19.5

98.4

11.7

40

22.6

131.2

13.5

50

25.2

164.0

15.1

60

27.6

196.8

16.6

70

29.9

229.6

17.9

80

31.9

262.4

19.2

90

33.9

295.2

20.3

100

35.7

328.0

21.4

1000

112.8

3280.0

67.7

2000

159.6

6560.0

95.7

5000

252.3

16,400.0

151.4

10,000

356.9

32,800.0

214.2

12,000

391.0

39,360.0

234.6

100,000

1,132.7

328,000.0

679.6

500,000

2,572.0

1,640,000.0

1543.2

1,000,000,000

1,006,344.9

3,280,000,000.0

603,806.9

Sanity check: if you are 0 meters off the surface of the Earth (lying down really really flat), the horizon is 0 kilometers away. That makes sense — you’re tangent to the surface! So the first line sounds right.

Now imagine you are standing on a beach, looking out over the ocean to the horizon. Most people aren’t two meters tall, and your eyes are several centimeters below the top of your head. But let’s just say your eyes are two meters off the ground (maybe you’re standing on a small sand dune). In that case, your horizon is 5.1 km (3 miles) away. That also sounds about right to me.

But now let’s say you are in your hotel overlooking the beach, and on your floor your eyes are 20 meters off the ground. The horizon is then 16 km away, much farther than before. Good: the higher you are, the farther away the horizon should be.

What if you’re a lot higher up, like in an airplane? At a cruising altitude of 39,000 feet (12,000 meters; typical for a cross-country flight) the horizon is 391 km (235 miles) away! That’s a surprisingly long way; in general that means you could be looking across one or more states in the US. This commonly fools me; seeing something even a little bit out from directly underneath the plane means it’s miles away.

What if you go up even higher? The Space Shuttle can reach a maximum height of about 500 km (actually a little more, but close enough). That’s 500,000 meters, or the second-to-last line of the table. For them, the horizon is almost 2600 km away! That means they can see almost the entire US by looking from one side of the Shuttle to the other. Cool.

And what if you’re really far away? From an infinite distance, you should see the horizon as being one Earth radius farther away than your height (draw a diagram if you want). In reality that’s impossible, so in the last line I put our poor observer floating in space one million kilometers away (more than twice the distance to the Moon). The horizon is then 1,006,344 km away, which is just about (but not quite) the Earth’s radius plus the observer’s distance over the surface. They are seeing almost — but not quite — half the Earth all at once.

So there you go. The next time you’re on a beach, or the next time you’re flying, take a look out to the horizon. Like the end of a rainbow, it’s impossible to reach. But it’s not impossible — it’s not even all that hard — to know how far away it is.

If you liked this, take a look at Mooey’s Top Ten ways to know the Earth isn’t flat. There’s even more geometric nerdity there. [Update: I had no idea, but Erik Rasmussen also has a writeup on this from last March, and it’s eerily similar to what I wrote. I swear I never saw his; but I guess great minds and all that!]

I sat atop Uluru (Ayers Rock) in the NT in central Oz once. It is about 350 metres high. Flat desert for miles all around except for the Kata Tjuta (The Olgas) a large rocky outcrop about 30km from The Rock. They look so close you feel like you could reach out and touch them. The horizon seemed miles away. Now, thanks to you BA, I know almost exactly how far away the horizon was. Too the south west was a range of mountain/hills on the horizon. I thought at the time that they had to be a range in South Australia which would put them more them almost 100 kay away. Wow.

That’s a cool calculation,
but the actual distance to the horizon is the arc length of the earth surface to the point, isn’t it?
I guess if you are in the surface of earth you are more interested in the arc length.
for a standing man the diference is very small but for a plane or the shuttle, I guess that the diference is reachable. =)

Nothing better than taking a couple minute break and checking into this blog for an article like this. And Southern Utah is great, but I got my life change (well, nudge really) flying over the Grand Canyon. Also, flying into Reagan National in DC is extraordinary.

SQR(height in feet) * 1.2 equals the distance to your horizon in miles.

This holds up very well to your mathematically derived method until you get to the extreme values. It has the advantage that it can be calculated pretty easily in your head, on the spot. Impress your friends!

I use the same kind of diagrams to show my friends that a full moon is actually impossible because the only time the conditions are right, is when the sun is behind the Earth, i.e. a lunar eclipse. In my case, it’s more “the moon can never look completely full.”

That’s right, isn’t it? I wouldn’t want to think I was telling people the wrong thing! It seems right to me, but no one of any authority has confirmed this.

Cool – this is really well explained. I often wanted to know how far I can see!

I live on the South Coast of England and if I stand on the shore line I can see ships on the Horizon in the English Channel. I wondered how far away they were – now, voilà!

Also If you stand on the cliffs at Dover (in SE England) you can just about see (if weather conditions are good) the coast of France. I estimate that the Cliffs are 200-300 meters in height and the coast of France is around 35-40km away – this proves the theory mathematically that you can see France from England!

I’d never thought about that before, guarantee it’s going to be rattling about my head the next time I’m flying. I will be fighting you for that aisle seat though – stretching the legs, easy access to the toilets and the stewards don’t have to reach as far to hand you that fourth gin and tonic…

Actually, this calculation is for a minimum distance. That’s because Earth’s atmosphere acts like a lens, bending the light rays. This atmospheric diffraction allows us to see somewhat farther than the geometric distance calculated here.

This calculation matters in other areas too – particularly line-of-sight radio propagation (VHF and higher frequencies) for amateur radio, FM broadcast, television, those little FRS sets and so forth.

At home, I have my VHF antenna on a 6m pole. That gives me a theoretical horizon of just shy of 9 km, modified by the local terrain to a receiver on the ground. Now nobody has a receiver sitting on the ground normally, but a car might have an antenna at 1m, and a home station might, like me, have an antenna at 6m. If you assume the line of sight from me to the car just touches the horizon, the maximum distance I could talk to a car is 8.7 km for me to the horizon plus 3.6 km for the horizon to the car for a total of 12.3 km. That is pretty close to what I see in practice, modified for the ability of radio waves to refract and diffract and be subject to a few other propagation tricks. OTOH, I regularly chat with someone who has an antenna up a 10m tower. His horizon is 11.3 km, which means he and I could chat if we’re within 20 km of each other, give or take. That is about the range between us (we’re actually a little farther apart, but radio waves can diffract over obstacles, just like light because, well, radio waves are light).

This is why antennas are often put on towers, of course, especially for FM broadcast. AM broadcast band needs towers because the antennas are very long (tens to hundreds of metres). FM broadcast band antenna elements are about1.5 metres long, give or take. But propagation at that frequency is line-of-sight so if they want a bigger area in which to sell advertising, they have to get that antenna up higher. Ditto for television stations.

One place you can use this… if you’re thinking of buying a pair of those little FRS radios to keep track of your kids at the cottage: The radios often come with wild claims of “19 km range”. Well, they’re UHF radios and are also line-of-sight. Realistically, you’re about 1.8m tall, and your child about 1.3 m tall – if the radio waves are just going to touch the horizon (that would give the maximum distance), you’re going to get ~5 km to the horizon from you and about ~4 km from the horizon to your child for a max range of ~9 km and that assumes there is NOTHING in the way – no trees, no grass, no cars, and certainly no hills or buildings. Realistically, it’s a LOT less than that, so don’t be fooled by the marketing claims on the box. Sure, you could use one of them from the ISS to the ground and talk 500 km, but that’s not really a useful comparison

This was excellent. To bolster your data. I work on the 24th floor of a skyscraper in Houston. On a very clear day, I can see just on the horizon a power plant that I know is exactly 25 miles away. If I assume 15 feet for every floor of the building then on the 24th floor I would be up 360 feet. Using your data 328 feet up would see 21 miles so it is just about right on.

Maybe someone will want to expand the formula that Phil gave to include not only the height of the observer, but the height of what is being observed as well. After all, don’t we want to know how far we can actually see (atmosphere notwithstanding) an object?

Bonus cool result: you can use exactly the same principles, but plugging in a known value for h and the angle to the horizon, to calculate the radius of the Earth, to a surpisingly good level of accuracy… Abū Rayḥān Muḥammad ibn Aḥmad al-Bīrūnī (973 – 1048) used this method to arrive at a figure of 6,339.9 km, which is only 16.8km out.

a lurker … that’s easy (in principle), use Phil’s calculation for your height, then assume your height is that of the object you want to observe (see how far you could see if you were at that height) and add the two together. However, as pointed out above, the distance you get will be the “straight” distance between your eyes and the top of the observed object (and not along arc).

p.s. I had to put “straight” into quotes, since the shortest path between two objects is such as light would take, which in gravitational field is “curved”.

p.s.2 I had to put “curved” into quotes, because technically it’s not the path that is curved, it’s really the space-time that’s curved – the path is actually a straight line in a curved space-time … 😀

A friend mentioned this to me a few months ago, and it all seemed counter-intuitive to me. Living in the Wyoming Valley in Northeastern Pennsylvania, surrounded by a ridge-and-valley mountain system, I am used to routinely seeing things much further away. From my house in Nanticoke I can see the lights on the ski slopes on Sno Mountain (formerly Montage Mountain), 25 miles away. I can also see wind turbines more than ten miles away. The concept of the “horizon” being at ground level has always escaped me. It’s useful in a flat desert or a calm body of water, I suppose.

Things got worse when I lived in the flat regions of Newark, Delaware, when the horizon was defined by the nearest buildings.

I copied and pasted that into a Word doc to save. Thanks Phil, that is very cool! I was always under the impression that you could see about 10 miles on the beach when the water was calm. Someone in college had mentioned that in one of my classes, and I just kind of took it as gospel. I should be more skeptical!!

Another way to use this lesson has to do with observing meteors. People see something bright flash low to the horizon, and think, “A few miles from here, just on the other side of that mountain.” In fact, the meteor is burning up at an altitude of 50-150 km, which means for it to appear near the horizon it’s really _hundreds_ of km distant.

Dittoing the trig formulation, although iirc I framed the calculation so it was the arctan instead of the arccos. There’s not much difference at small elevations, but you really want the distance on the ground, not the LOS distance, in most cases.

As it happens this gives you a good introduction to approximations for the kiddies. cos(t) ~ tan(t) ~ t ~ arccos(t) ~ arctan(t) for small values of ‘t’, something that is definitely true in the case of low elevations. I ended up with an approximation of sqrt(2Rh) for the distance to the horizon (again, iirc). Add in the second term and you get a better approximation, and for grins you can calculate the difference to see the range where the first approximation is ‘good enough’.

There are two other interesting things you can do with this — first, figure out what you can see out of your plane window. An entire Great Lake? No problem. Florida from coast to coast? Ditto. (This doesn’t work well in practice due to sea haze.) In the center of the country you just see plains, but it can be interesting on the edges.

Second, figure out how far you can see storms from the mountains. I remember reading a Discover(?) story on sprites, imps and similar phenomena, and the guy mentioned doing research from his place in the mountains near Denver. He said he could see storms from Texas to South (North?) Dakota. At first I thought that was a wild exaggeration, but then I figured out the distance to his horizon, plus the elevation of the top of major thunderstorms, and realized that he was right. You can probably see storms 300 miles away if you’re near 10k feet, with the limiting factor atmospheric haze. You could see sprints from much further away.

@Shane
It only works when you assume flat, sea-level land/water. When you start including peaks and views from peaks then you need a specific equation.

In the Rockies, we can stand on a high peak and see peaks well beyond the normal horizon. Also, you have to account for the altitude of the surrounding land. For example, when you stand on Pikes Peak at 14109′, Colorado Springs, at your feet, it over 6000′, so you’re not 14,000′ above the plane of vision; yet to the east, the plains are descending in elevation all the way to the eastern horizon, changing the horizon point (not by much if I once calculated it correctly). Similarly, Denali (at 20,320′ and the highest-relief mountain from the surrounding plains) has been spotted from 300 hundred miles away, well “beyond its horizon”.
(also with nods to what Andre says above)

# Jonathan Says:
“What about the bluing of objects at a distance? Wouldn’t this effect prevent you from seeing very very far away?”
It doesn’t make objects disappear (though in practice certain atmospheric conditions can make things disappear), it just makes them look blue because that’s the color wavelength that travels farthest. The object is still visible but the other colors wavelengths have “worn out and given up”.

Actually, Phil is wrong in his last paragraph. It /is/ possible to reach the horizon. It’s not even all that difficult a trick: Don’t go chasing it, let /it/ chase /you/, as you lower yourself until your eyes reach that magic height of 0.

Last fall, I stood on a beach at night, looking off across a bay towards a distant navigation channel. It’s in a remote area, so there were no light pollutants.

A couple miles out, I could see one of the navaid buoys, whose lamps bobs about what, 6 or 8 feet above the surface of the water.

About twice that distance out, I could just barely make out another. Sometimes it seemed to disappear for one or two pulses, and then come back. And then I realized that it might be the bobbing of the buoy, or waves at the very edge of the horizon, that made it visible at times, and hidden at others.

I slowly backed up the beach, a difference in elevation of maybe three feet at most… and voilà! there was the second buoy, blinking as regularly as the first.

So I got a little higher, on the top of the small dune, maybe six or seven feet above the edge of the water. And from there, I could see other navaids, even further out.

So I walked back down to the water’s edge, and backwards back up to the dune, and back, and forth, and back, and forth, watching the lights appear and disappear. At times, I could even make the far lights appear and disappear from view on cue, just by standing on tiptoes, or scrunching down my shoulders.

I’ve seen the altitude effect on horizon distance over the same bay many times; as you climb the hill in town, more and more details of the shore on the opposite side come into view. And there’s a radio tower strobe light that you can see from one part of the beach, but not another, because your are “moving” an intervening hill in and out of the way. But this was the first time I’ve ever observed the curvature of the earth, a PLANET, so immediately. It was a question of INCHES.

Normally I’m not into maths, but this was a really cool post. Easy to understand, and more importantly, explains something we don’t even think about. I know now that the horizon outside my window is about 5km away – and I should have already known that, since I drive about that far to get to the local shopping centre on the horizon!

A possibly simpler way to do the calculation is with Euclid’s Book III Proposition 36, which states that the square of the tangent to a circle from an external point is the product of the two distances from the point to the circle along any line. Just let the line pass through the circle’s center and you get the same formula.

@Jeffersonian, that makes sense. We’re only talking peaks of a few hundred metres or so and when I did a quick map measurement the hills were further away than the formula says. I suppose a quick rule of thumb would be that hills and mountains are further away than you think.

If a person of average hight stands a little above the water line, the horizon will be 3 miles away. Hence, I take it, the traditionasl ‘Three mile limit’ for off shore soverenty. Not, as often said, the range of costal batteries.

this isn’t quite correct, the earth is an ellipse, not a circle. It’s rotation makes it slightly wider at the equator than it is between the two poles. The real calculation would be a lot more complicated.

Well, these calculations are wrong. Refraction must be taken into account. I know beacuse I learned how to calculate horizon distance for radio propagation when I was at University and the effect of refraction was quite significant. This means that the horizon is actually farther, and I insist, this effect is quite important, it ‘moves’ the horizon several kilometers.

Refraction bends the light path down, and lets you see a little further than you could if there were no air. You can account for refraction pretty well by using your formula, but substituting an effective earth radius that is 4/3 times the actual radius. This “four-thirds earth” approximation is common in the radar community.

I remember seeing this exact formula in GURPS Mecha, so you could tell whether a mech (whether by being taller or by flying) could target another with direct-fire weapons like lasers or such. But then again, this is GURPS, and they love calculations (you use cube roots to help get the size of your designed mecha).

I remember when I started taking flying lessons, I had problems determining distance by sight because of just how far I could see. It really is amazing. Quite a thing to get used to the idea of “The next town is 40 minutes away” and then being able to see it from my window while still over home.

The world is still pretty from up above. More people should fly so they can see it. It’s even better with small airplanes because you can get low enough to get a good look at things and slow enough to have time to do so. When they say “Free as a bird” they aren’t kidding. Go to your local small airport and find an instructor, and give him $50 or whatever it costs for the “intro ride”. You’ll love it. Trust me. Just be forewarned that the flying bug bites HARD.

A good old seaman trick…
The distance to the horizon expressed in NAUTICAL MILES is at:
d=2*sqrt(h)
when h is expressed in meters.
For example, if you’re standing at the deck of a cruise ship (about 30 meters above the water), the horizon will be at approximately 11 nautical miles or 12.6 miles (give or take, nautical miles + 15%).

I had never previously calcuated the distance of the horizon (while standing) but I decided to make an educated guess. My guess was 3-5miles. After reading your article, I was apparently closer than I expeced.

Most people seem to like this problem. Maybe you like this one to:
Travelling from New York City to Washington DC in the shortest trajectory, that is through the earth, (not over its surface!) you decide halfway that you want a refreshment. How far do you have to dig up?

Respectfully, your calculations are inaccurate. While they may (or may not, I did not proof them) stand up mathematically in reality they do not.

I live in Tampa, FL USA.

Standing at 5’10’ at sea level from one end of the Skyway Bridge I can see the other side where it meets land. According to Wikipedia and my vehicle’s odometer the bridge is approximately 5.5 miles across. This is pretty much twice the distance you describe as the distance to visible horizon as my eyes are at approximately 5’8″.

On Maui I watch the sun illumination on the opposite mountain move down in the morning. This suggests a little extra credit calculation you folks might enjoy doing. After sunset you’ll see clouds above you (or north or south of you) still illuminated. By measuring the time from sunset on you to sunset on the clouds, you’ll be able to compute their height. Enjoy!

Coincidentally on the west coast we often watch for the “green flash”, a phenomena sometimes observable on the horizon at sunset when the air is dry enough. One day looking down to the beach from a small cliff (about 100′), it was clear the people below saw it before we did because they turned attention away from the sunset. Although the sun leaves a wide fuzzy shadow, the precise time it disappears from view is easily noted within a few seconds.

Absolutely the most beautiful thing I have ever seen in my life had to do with the horizon.

I was on an overnight flight and woke up to look out the window and saw the sun coming up just below the horizon with the stars in a perfectly clear sky above – it was just at that point where the sunlight wasn’t strong enough to overwhelm the starlight, but was there enough to give color to the haze.

Yes, it’s a really simple calculation that I’ve been using by the time of our first test in my high school physics classes for a while. And if you let 2r= D, the diameter of the planet, then for small heights (h<<r), it's just SQRT (Dh). At that point, you can show the kids that just by dimensional analysis, the answer has to be at least proportional to that.

I bring it up by telling the kids that 6-foot tall Buzz Aldrin said that when he was on the Moon, he had the impression that he was on a ball. And that's because the horizon was only about 1.5 miles away from him. On Earth, it would be 3 miles.

One has to realise that this calculation is not new, and is in fact of primary interest to seafarers wondering how far away that lighthouse is. There is a series of documents called Admiralty List of Lights and Fog Signals, published in the UK by the Hydrographic Office. It comes in 11 volumes, each covering a specific part of the world. Being a parochial Brit, I’ve only ever seen Volume A, which is my local coastline, but I expect that Volumes J and K, which deal with the American coastlines, are similar.

Volumes can be bought via the Net (ABE books, for example) for about half the cover price.

Now the point is that Page ii of each Volume is a Geographical Range Table. It tells you how far away things are for a height of eye between 3 and 150 feet, and for an elevation of the thing you’re looking at from 0 to 1300 feet. The 0 feet line therefore shows the horizon distance, and the table as a whole answers the question raised above about adding the two ideas together. The distances calculated take account of refraction, as well. But notice, all the answers are in sea miles.

WJM,
mariners called that “bobbing the light”. If they spotted a lighthouse of known height, they could go up and down ladders to change their height of eye, find their height of eye when the light would just disappear, and thereby deterimne the distance to the lighthouse. I did it a few times on a submarine I was on. I had the OOD slowly change depth and got the distance to one of the lights on San Clemente island.

Tom’n’Effie,
the U.S. version is called “Light Lists” and the U.S. version of other countries is called the “List of Lights”. Google “NGA light list” and you’ll find em.

Distance to the horizon (nautical miles) = sqrt(height of eye(ft))*1.17. This formula accounts for atmospheric refraction. For the radar horizon (for common marine band radars), change the constant to 1.23.

i dont know anything about calculations, but on a clear day in brighton (u.k) you can stand by the pier and look down the coast and see worthing pier and its a good 14 miles away. 3 miles? you can see france from dover. FRANCE. with your eyes. from ENGLAND. i can see how height can affect your vision but there isnt 50 metres of sand on any beach ive ever been on.

It has been nearly four days’ now since Dr. Phil Plait posted this article, and I’m surprised to find that Tom Marking still hasn’t showed up on this thread with his usual long list of equations and data… yet!

One year I thought I was going to have to miss the Perseid Meteor Shower because I was going out of town. I turned out that my window seat from Atlanta to Boston happened to be the front row seat for viewing it!

I was above the haze, clouds, and light pollution — and I had nothing better to do. I pulled the navy Delta Airlines blanket over my head and peered out the window. There was a ‘boy scout’ looking little boy in the row in front of me who asked what I was doing. He also got a blanket and we “ooh”-ed and “ahh”-ed the whole way.

Phil, I always enjoy your posts and this one’s no different, but I’ve got to get on your case for the “Amurcan” crap. I spend a lot of time online and in any discussion this kind of thing pops up. I don’t think you’re using it in an intentionally insulting way here but it is anyway. “You were born in a particular spot on the globe, therefore you’re too dumb to speak correctly / figure out a chart / etc…”
I know it’s a cliche and it shouldn’t bother me, but as I see it, you should be above that kind of thing.
Everyone already hit all of the math permutations, so I had to post something different.

I always wondered if the horizon is the result of the curvature of the earth or the limitations of our eyes.

Isn’t perspective causing things to shrink the further they are away. Imagine a long straight lane with lamp posts next to it, each of the same height. You will see that each lamp posts gets smaller than the one in front of it. This causes our feeling of perspective. Now, could the horizon simply be the location where these lampposts become so small, we can’t see them anymore?

Forget the maths, go to the beach and have a look. I can stand on the beach at Fremantle in Western Australia and clearly see Rottnest Island that I know is 12 miles away. I can see the beaches and the boats in the bay with my binoculars. When ships sail past Rottnest, heading towards India, I can see them sailing away for over 1/2 hour. I suggest the horizon, while standing at sea level and from eye height(approx 5 ft) in more like 25 miles, not 3 miles like all the scientist appear to have calculated.

I have always enjoyed this concept, even though I do not have a scientific mind whatsoever, I do appreciate the formula above…very interesting. One example of this topic, outside of watching the horizon line from inside an airplane, was standing on top of Pikes Peak and seeing downtown Denver to the north and Raton Pass to the south, well over a 200 mile spread. Thanks again for the info…

I thought I was asking a simple question! Wow, I didn’t know it was that complicated. No wonder everyone I asked couldn’t come up with the answer.Thanks for all the info.Although it was way more than I needed or could understand.

I thought I was asking a simple question! Wow, I didn’t realize it was so complicated to figure out.No wonder everyone I asked didn’t have a clue.Thanks for the info. Although it was way more than I needed. And way more than I could understand.

I was standing on the shoreline in Huntington Beach, Ca and looking out over the ocean at Catalina Island at least 20 miles away. I think I saw pretty close to the bottom of the island. How’s that possible if I should not be able to see the horizon past 5 or 6 miles?

There is a simpler way to compute the horizon distance: let Γ be a circle of center O and radius r and A any point. Then, for any line Δ through A crossing Γ at the two points M and M’, we have AM.AM’ = OA²-r²; this number is called the power of A relatively to Γ, and note that it does not depend on the line Δ.

In the present case, A = you, Γ = Earth, and you can compute the power by using two different lines Δ, namely the tangent line and the vertical line; this directly yields

(distance to horizon)² = (altitude).(diameter of Earth + altitude)

or, if you are in a plane or lower (and thus altitude is negligible compared to the diameter of the Earth),

(distance to horizon) = √(altitude.diameter of Earth)

Thus, at low altitudes, when going four times higher, you see twice further.

Of course, this is assuming that the Earth is a perfect sphere. When you are on the beach, you are probably actually a bit above sea level (on a jetty) and looking at an object that itself is above sea level (island, ship mast). To see that object, there must be some horizon that both you and the target are able to see, and thus you need to add the horizon distances for you and the target. For example, one pirate ship could see (from the crow’s nest, 30m above sea level) up to 20 km away, thus two such ships would see each other from 40 km apart.

And as far as window seats go, my favourite one was flying over (and actually, around) the western Himalaya range, including the conflictual zones of Pakistan and Afghanistan. Man, those mountains in the background are *high*.

This is very interesting indeed. Although I don’t really have a scientific or even mathematical mind, I’m starting to doubt my cousin’s contention that the land is so flat around his Wyoming home, that he can watch his dog run away for 3 days.
He is either pulling my leg, or he has a very tall dog.

I accidentally happened onto this website and found it very interesting since I, too, have arrived at a formula to find the distance to the horizon from a certain altitude above a given sphere.

It is a very accurate and quite simple formula when compared to those shown above. It doesn’t get any simpler than this.

It reads: distance to the horizon equals the square root of altitude (a) times (diameter of the sphere (s) plus altitude (a)):

d=sqrt|(a(s+a)|

a= altitude above sphere
s= diameter of sphere
d= distance to horizon

For example:

At an altitude(a)of 1 mile above Earth(s),7910 miles in diameter,the distance(d)to the horizon would be 88.9438 miles.
or
At an altitude(a)of 1 mile above the Moon(s),2160 miles in diameter, the distance(d)to the horizon would be 46.4865 miles.
or
At an altitude(a) of 1 mile above the Sun(s),864,000 miles in diameter, the distance(d)to the horizon would be 929.5165 miles.

This formula, in its simplicity, can be used with any unit of measurement, size of sphere or altitude, however small or large.

Need help with a similar but slightly more complex question. If an object is 100 feet tall out on the ocean and I am 10 feet above the waters surface, approximately how far away can it been seen? Basically I am trying to figure out if an oil plat form can be easily seen 10 miles out from the shore

30. Harold Says:
January 15th, 2009 at 10:55 am
…The concept of the “horizon” being at ground level has always escaped me….

What this mean is, if an observer is looking straight out and level (that is, at a 90 degree angle to a line from his feet to the center of the Earth, the visible horizon will always appear to be at his own eye level. You can check this easily by standing on the beach and looking out level, then going to the top of a tall building and looking out level, and then go up in a jet liner and looking out. Sure enough, the horizon line will seem to rise up with you, at least up to jetliner height – 30,000 feet.

Now here’s my question. If you were to go high enough, up to the level of the space station, say, it seems like the horizon would eventually have to drop below your eye line – I mean, the Earth’s only so big.

Has anyone out there (an astronaut or a U2 pilot?) seen if this happens? And if it does, how high do you have to go before it kicks in?

This has been bothering me for years, which will give you an idea of what a quiet life I lead…. answers would be appreciated.

So, if I understand Bob and Jerome correctly, would that not mean that if I want to find the maximum distance at which an item of altitude A can see an item of altitude B that I would do maxDistance = sqrt(altA * (d * altA)) + sqrt(altB * (d * altB)) ? I think that is the answer for Bill’s oil-rig question.

There’s also the other simple quick approximation
.#miles= sqrt(1.5*height in feet)
which a fairly accurate because the ratio of the Earth’s diameter in miles to the number of feet in a mile is very close to 3/2.